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VCE 11 General 2023

10.03 Perimeter

Lesson
Perimeter

The perimeter is the distance around a two dimensional shape.

To find the perimeter of any polygon (straight sided shape) add all the lengths of the sides, ensuring all lengths are written in the same unit.

Perimeter is a measure of length, so different units of measure could be used to measure the perimeter of different sized objects. Millimetres could be used to measure the perimeter of a SIM card, centimetres to measure the perimeter of a wallet, metres to measure the perimeter of a room and kilometres to measure the perimeter of a town.

If no particular unit for the context or question is given it is also good mathematical practice to use the word 'units'.

 

Calculating Perimeters

Triangle

Here is a scalene triangle. To find the perimeter add the three side lengths. Notice that all sides are measured using the same units.

Perimeter $=$= $14+12+6$14+12+6
  $=$= $32$32 cm

 

Quadrilaterals

Here is a rectangle. Recall that rectangles have opposite sides of equal length.

Perimeter $=$= $2\times13+2\times6$2×13+2×6

(there are two $13$13 mm sides and two $6$6 mm sides)

  $=$= $26+12$26+12

 

  $=$= $38$38 mm

 


A square has $4$4 sides of the same length, so the perimeter of a square will be $4$4 times one of the side lengths.

Perimeter $=$= $4\times7.4$4×7.4

(there are four $7.4$7.4 cm sides)

  $=$= $4\times7+4\times0.4$4×7+4×0.4

 

  $=$= $28+1.6$28+1.6

 

  $=$= $29.6$29.6 cm

 

 

Irregular shapes

All perimeters can be found by traveling around the shape, adding up one side at a time.

Here is a trapezium. To find its perimeter add the side lengths.

Perimeter $=$= $1.3+1.2+2.7+2.4$1.3+1.2+2.7+2.4
  $=$= $7.6$7.6 m

 

Also keep in mind the ability to construct simple rules for other shapes as necessary, and to group together sides of the same lengths.

Perimeter $=$= $3\times1+2\times3.6+1.4+1.5+3.45$3×1+2×3.6+1.4+1.5+3.45
  $=$= $3+7.2+2.9+3.45$3+7.2+2.9+3.45
  $=$= $10.2+6.35$10.2+6.35
  $=$= $16.55$16.55 m

 

Practice questions

Question 1

Find the perimeter of the shaded region shown, where edges are 1 unit in length each.

  1. $\editable{}$ units

Question 2

Find the perimeter of the figure shown.

 

Circumference

The perimeter of a circle, the circumference, can be used to find the perimeter of composite shapes that involve full circles, or semicircles.

Circumference

The circumference $C$C of a circle is given by:

$C=2\pi r=\pi d$C=2πr=πd

where $r$r is the radius and $d$d is the diameter of the circle.

This is a composite shape, made up of a semicircle and a rectangle. Although, we are missing one side of the rectangle, we can use the circumference of a circle, and halve this, to find the length of the rounded edge.

We have $3$3 straight sides, two with length $4$4 cm and one with length $8.4$8.4 cm. The sum of these lengths is equal to:

$2\times4+8.4=16.4$2×4+8.4=16.4 cm

Lastly we have a semicircle of radius $\frac{8.4}{2}=4.2$8.42=4.2 cm. So the length of the rounded edge is equal to:

$\frac{2\pi\times4.2}{2}=4.2\pi$2π×4.22=4.2π cm.

In total, the perimeter is:

Perimeter $=16.4+4.2\pi\approx29.6$=16.4+4.2π29.6 cm ($1$1 d.p.)

 

Practice questions

Question 3

Find the circumference of the circle shown, correct to two decimal places.

A circle with a radius line drawn from the center to the circumference, labeled as 8 cm. An arrowhead points to the end of the radius line on the circumference.

Question 4

Find the perimeter of the trapezium shown.

Question 5

Find the perimeter of the shape (shaded) shown.

  1. Give your answer correct to $2$2 decimal places.

 

Arc lengths and sectors

An arc of a circle is a curved line formed from part of the circumference of the circle. The length of an arc is called the arc length, $l$l.

sector is a shaped formed from part of a circle, where the sector's boundary or perimeter is formed by two radii and an arc.

Finding the perimeter of a sector involves first calculating the arc length, then adding on the lengths of the two radii.

Examples of sectors

 

Note that a quarter of a circle, is sometimes called a quadrant, and half of a circle is usually called a semicircle.

 

Worked example

Example 1

Find the perimeter of the sector equal to $\frac{1}{4}$14 of a circle with radius $6.5$6.5 cm.

Think: To find the perimeter of the sector, we start by finding the arc length, $l$l. This is equal to $\frac{1}{4}$14 of the circumference of the circle. 

Do:

arc length $=$= $\frac{1}{4}\times\text{circumference }$14×circumference

 

$l$l $=$= $\frac{1}{4}\times2\pi r$14×2πr

 

  $=$= $\frac{1}{4}\times2\times\pi\times6.5$14×2×π×6.5

(Substitute $r=6.5$r=6.5)

  $=$= $3.25\pi$3.25π

(Leave answer in terms of $\pi$π)

 

If we leave our answer for arc length in terms of pi, we eliminate any rounding error in our calculations.

To get the perimeter, we must add the lengths of the two radii to the arc length.

perimeter $=$= $\text{arc length }+2\times\text{radius }$arc length +2×radius

 

  $=$= $3.25\pi+2\times6.5$3.25π+2×6.5

(Substitute $r=6.5$r=6.5)

  $=$= $3.25\pi+13$3.25π+13

 

  $=$= $23.2101$23.2101...

 

  $=$= $23.2$23.2 cm ($1$1 d.p.)

(Rounded to $1$1 decimal place)

 

Practice questions

Question 6

What is the perimeter of a semicircle with diameter $8$8 cm, correct to two decimal places?

Question 7

A circle with radius $4$4 cm has been drawn with a dashed line.

A sector is outlined with a filled line.

  1. Find the exact circumference of the whole circle.

  2. Find the exact length of the arc of the sector.

  3. Find the perimeter of the sector.

    Round your answer to two decimal places.

Outcomes

U2.AoS4.5

the perimeter and areas of triangles (using several methods based on information available), quadrilaterals, circles and composite shapes, including arcs

U2.AoS4.11

calculate the perimeter and areas of triangles (calculating the areas of triangles in practical situations using the rules A=1/2 bh, A=1/2 ab sin(c) or A=\sqrt{s(s-a)(s-b)(s-c)} where s=(a+b+c)/2

U2.AoS4.12

use quadrilaterals, circles and composite shapes including arcs and sectors in practical situations

U2.AoS4.13

calculate the perimeter, areas, volumes and surface areas of solids (spheres, cylinders, pyramids and prisms and composite objects) in practical situations, including simple uses of Pythagoras’ in three dimensions

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